Given: \[ x[n] = e^{j\omega_0 n} (0.5)^{|n|} \] We are to find the **Discrete Time Fourier Transform (DTFT)** of \( x[n] \). --- ## **Step 1: Write the DTFT Definition** The DTFT of \( x[n] \) is: \[ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \] Substitute \( x[n] \): \[ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} e^{j\omega_0 n} (0.5)^{|n|} e^{-j\omega n} \] \[ = \sum_{n=-\infty}^{\infty} (0.5)^{|n|} e^{j(\omega_0 - \omega) n} \] --- ## **Step 2: Split the Sum for \( n \geq 0 \) and \( n < 0 \)** Recall: - For \( n \geq 0 \): \( |n| = n \) - For \( n < 0 \): \( |n| = -n \) So, \[ X(e^{j\omega}) = \sum_{n=0}^{\infty} (0.5)^{n} e^{j(\omega_0 - \omega) n} + \sum_{n=-\infty}^{-1} (0.5)^{-n} e^{j(\omega_0 - \omega) n} \] For the second sum, change variable \( m = -n \), so as \( n \) goes from \( -\infty \) to \( -1 \), \( m \) goes from \( \infty \) to \( 1 \): \[ \sum_{n=-\infty}^{-1} (0.5)^{-n} e^{j(\omega_0 - \omega) n} = \sum_{m=1}^{\infty} (0.5)^{m} e^{-j(\omega_0 - \omega) m} \] --- ## **Step 3: Express the Whole Sum** So, \[ X(e^{j\omega}) = \sum_{n=0}^{\infty} (0.5)^{n} e^{j(\omega_0 - \omega) n} + \sum_{m=1}^{\infty} (0.5)^{m} e^{-j(\omega_0 - \omega) m} \] Let’s write both sums with the same index: \[ X(e^{j\omega}) = \sum_{n=0}^{\infty} (0.5)^{n} e^{j(\omega_0 - \omega) n} + \sum_{n=1}^{\infty} (0.5)^{n} e^{-j(\omega_0 - \omega) n} \] --- ## **Step 4: Factor and Recognize Geometric Series** Let \( a = 0.5 e^{j(\omega_0 - \omega)} \), \( b = 0.5 e^{-j(\omega_0 - \omega)} \): \[ X(e^{j\omega}) = \sum_{n=0}^{\infty} a^{n} + \sum_{n=1}^{\infty} b^{n} \] \[ = \sum_{n=0}^{\infty} a^n + \left( \sum_{n=0}^{\infty} b^n - b^0 \right ) \] But \( b^0 = 1 \): \[ = \sum_{n=0}^{\infty} a^n + \sum_{n=0}^{\infty} b^n - 1 \] --- ## **Step 5: Compute the Geometric Sums** Recall for \( |r|<1 \): \[ \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} \] So, \[ \sum_{n=0}^{\infty} a^n = \frac{1}{1-a} \] \[ \sum_{n=0}^{\infty} b^n = \frac{1}{1-b} \] So, \[ X(e^{j\omega}) = \frac{1}{1 - a} + \frac{1}{1 - b} - 1 \] Recall \( a = 0.5 e^{j(\omega_0 - \omega)} \), \( b = 0.5 e^{-j(\omega_0 - \omega)} \): \[ X(e^{j\omega}) = \frac{1}{1 - 0.5 e^{j(\omega_0 - \omega)}} + \frac{1}{1 - 0.5 e^{-j(\omega_0 - \omega)}} - 1 \] --- ## **Step 6: Final Answer (Boxed)** \[ \boxed{ X(e^{j\omega}) = \frac{1}{1 - 0.5 e^{j(\omega_0 - \omega)}} + \frac{1}{1 - 0.5 e^{-j(\omega_0 - \omega)}} - 1 } \] --- ### **Summary of Steps** 1. Substitute the given signal into the DTFT definition. 2. Split the sum into \( n \geq 0 \) and \( n < 0 \). 3. Change variables in the negative sum to positive limits. 4. Recognize and sum the geometric series. 5. Substitute back and simplify. --- ### **If you want to further simplify:** You can combine the two fractions for further manipulation, but the above boxed answer is the standard, compact form. --- **If \( \omega_0 = 0 \), this reduces to:** \[ X(e^{j\omega}) = \frac{1}{1 - 0.5 e^{-j\omega}} + \frac{1}{1 - 0.5 e^{j\omega}} - 1 \]